Back to the proof. Let ω be an n-ary operator on A and (a1,…,an)≡(b1,…,bn)(modΘ). We want to show that ω⁢(a1,…,an)≡ω⁢(b1,…,bn)(modΘ).

The proof now breaks up into two main steps:

Step 1.

For each i∈n, ai≡bi(modΘ) means we have a finite

ai=c1≡F1c2≡F2⋯≡Fp-1cp=bi,

where each Fi is a congruence in C, and each ci∈A. Now the lengths of the sequences may vary by i. The idea is to show that we can in fact pick the sequences so that they all have the same length.

To see this, take the first two pairs of congruentelementsa1≡b1(modΘ) and a2≡b2(modΘ), and expand them into two finite

1.

a1=c1≡F1c2≡F2⋯≡Fp-1cp=b1, and

2.

a2=d1≡G1d2≡G2⋯≡Gq-1dq=b2,

where ci,dj∈A and Fi,Gj∈𝒞. If q<p, we may lengthen the second chain so it has the same length as the first one:

a2=d1≡G1d2⁢⋯≡Gq-1dq≡Gqdq+1≡Gq+1⋯≡Gp-1dp,

where Gq-1=Gq=⋯=Gp-1 and dq=⋯=dp=b.

The above argument and an induction step on n show that we can indeed make all the “expanded” the same length. As a result, without loss of generality, we assume that all the expanded chains have the same length, say r+1.

Step 2.

Instead of writing all n chains, let us use our notational device, and we have the following single (again, we may write it in this fashion because all chains are now assumed to have the same finite length of r+1):

But this chain implies that ω⁢(a1,…,an)=ω⁢(c11,…,ci⁢n)≡ω⁢(c21,…,c2⁢n)(modΘ). So what we have shown is that the images of the first congruence pair are congruent modulo Θ. But this process can be easily applied to subsequent congruence pairs, so that we end up with

Conversely, it can be shown (by Grätzer) that every algebraic lattice is isomorphic to the congruence lattice of some algebraic system.

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If A is a lattice, then Con⁡(A) is distributive. The converse statement, that every distributive algebraic lattice is isomorphic to a congruence lattice, has recently been proven to be false by F. Wehrung.